therefore I understand that if you role a typical pair of dice, your possibilities of obtaining Snake eyes (double 1s) is \$1\$ in \$36\$. What I"m not certain of is how to carry out the civicpride-kusatsu.net to figure out your opportunities of rolling line Eyes at least once throughout a collection of rolls. I understand if I roll the dice \$36\$ time it won"t cause a \$100\%\$ possibility of rolling snake Eyes, and also while i imagine it"s in the top nineties, I"d favor to figure out exactly how i can not qualify it is.

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The probability the hitting that at the very least once is \$1\$ minus the probabilty that never hitting it.

Every time you roll the dice, you have actually a \$35/36\$ possibility of not hitting it. If you role the dice \$n\$ times, climate the only case where you have actually never hit it, is when you have not hit that every single time.

The probabilty of not hitting v \$2\$ roll is for this reason \$35/36 imes 35/36\$, the probabilty of no hitting with \$3\$ rolfes is \$35/36 imes 35/36 imes 35/36=(35/36)^3\$ and also so ~ above till \$(35/36)^n\$.

Thus the probability that hitting it at least once is \$1-(35/36)^n\$ where \$n\$ is the number of throws.

After \$164\$ throws, the probability of hitting that at the very least once is \$99\%\$

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edited Oct 4 "18 at 13:00

J. M. Ain't a civicpride-kusatsu.netematician
reply Oct 3 "18 at 13:19

b00n heTb00n heT
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The other answers explain the general formula because that the probability of never ever rolling snake eyes in a series of \$n\$ rolls.

However, you likewise ask specifically about the instance \$n=36\$, i.e. If you have actually a \$1\$ in \$k\$ chance of success, what is your opportunity of obtaining at least one success in \$k\$ trials? It transforms out the the answer come this inquiry is quite similar for any kind of reasonably huge value that \$k\$.

It is \$1-ig(1-frac1kig)^k\$, and \$ig(1-frac1kig)^k\$ converges come \$e^-1\$. Therefore the probability will be about \$1-e^-1approx 63.2\%\$, and this approximation will certainly get better the bigger \$k\$ is. (For \$k=36\$ the actual answer is \$63.7\%\$.)

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answer Oct 3 "18 at 13:40

especially LimeEspecially Lime
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If you roll \$n\$ times, climate the probability of rolling line eyes at least once is \$1-left(frac3536 ight)^n\$, together you either roll snake eye at the very least once or not at all (so the probability of this two events should sum to \$1\$), and also the probability of never rolling line eyes is the same as requiring that you roll among the various other \$35\$ possible outcomes on every roll.

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reply Oct 3 "18 in ~ 13:22
Sam StreeterSam Streeter
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